A force F, with a magnitude of 193,19 N, is applied at an angle of 20° with the horizontal to a block of mass 55 kg - NSC Technical Sciences - Question 2 - 2024 - Paper 1

Newton's First Law of Motion states that an object continues in a state of rest or uniform motion (moving with a constant velocity) unless it is acted upon by a net external or unbalanced force.

Since the block moves at a constant velocity, the net horizontal force acting on the block is 0 N. This can be expressed using the equation: $F_{net} = 0$ Where $F_{net}$ is the sum of all forces acting in the horizontal direction.

The vertical component of the applied force $F$ can be calculated using the formula: $F_{y} = F imes ext{sin}( heta)$ Where $F = 193.19 ext{ N}$ and $heta = 20°$. Thus: $F_{y} = 193.19 imes ext{sin}(20°) = 66.07 ext{ N}$

The normal force can be found by considering the forces acting vertically. It can be calculated as: $F_N = m imes g - F_y$ Where $m = 55 ext{ kg}$ and $g = 9.81 ext{ m/s}^2$: $F_N = 55 imes 9.81 - 66.07 = 605.07 ext{ N}$

The frictional force ($f_k$) can be calculated using the formula: $f_k = ext{friction coefficient} imes F_N$ Given the coefficient of kinetic friction is 0.3, we have: $f_k = 0.3 imes 605.07 = 181.52 ext{ N}$

When the force is applied at an angle of 0°, the entire force acts horizontally, reducing the vertical downward force caused by gravity that the normal force must counteract. Thus, the normal force decreases because there is no vertical component of the applied force acting upwards to subtract from the gravitational force.